From the acceleration-velocity graph, we have $\mathrm{a}=\mathrm{k}\mathrm{ }\mathrm{v}\mathrm{ }$where $\mathrm{k}$ is a constant which represents the slope of the given line.
As, $\mathrm{a}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{ds}},$ so $\mathrm{v}\frac{\mathrm{dv}}{\mathrm{ds}}=\mathrm{kv}$
or $\frac{\mathrm{dv}}{\mathrm{ds}}=\mathrm{k}=\mathrm{a}$constant
Thus, the slope of the velocity-displacement graph is the same as that of acceleration-velocity. Which is constant.